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Consider the differential operator $T:u\mapsto -iu'$ for any $u\in D(T):=\{f\in AC[-\pi,\pi]~|~f'\in L^2(-\pi,\pi),f(-\pi)=f(\pi)\}$; we consider $T$ as a densely-defined operator on $L^2(-\pi,\pi)$. It is easily seen, using integration by parts, that $T\subseteq T^*$, i.e., $T$ is symmetric. I wish to show that $T$ is in fact self-adjoint. What would be the best route to achieving this result?

It is clear from symmetricity of $T$ that $T^{**}\subseteq T^*$. Does symmetricity of $T$ also imply that $T^*$ is symmetric? This would mean that $T^*=T^{**}$. If this were the case then a sufficient condition for self-adjointness of $T$ would be closedness of $T$ (since $\overline{T}=T^{**}$).

Any help on this would be greatly appreciated!

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  • $\begingroup$ On what Hilbert space are you defining $T$? $AC[-\pi,\pi]$? $\endgroup$ – Tom Nov 30 '13 at 18:09
  • $\begingroup$ $L^2(-\pi,\pi)$; edited the question correspondingly. $\endgroup$ – Stromael Nov 30 '13 at 18:31
  • $\begingroup$ Any ideas, please? :( $\endgroup$ – Stromael Dec 2 '13 at 10:33

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